Tentukan titik stasioner dari persamaan f(x)= (x^2-1)/(x+2)

jawab

y = (x² -1) /(x+2)
y = u/v → y' = (u' v - uv')/(v²)
y'= 0

y' = (x² + 4x +1) /(x+2)²
y' = 0 → x² + 4x + 1= 0
(x + 2)² = -1 + 4 =  3
x + 2 =  +_√3
x₁ = -2 + √3
x₂ = - 2 - √3

y = (x² -1)/(x+2)
x₁ = -2 + √3  → y₁ = {(-2 +√3)² - 1 } /(-2 +√3) + 2)
y₁ = (4+3 - 4√3 - 1) / (√3)
y₁ = (6 - 4√3) /(√3)
y₁ = 1/3 (6√3 - 12) = 2√3 - 4

x₂ = -2 - √3 → y₂ = -2√3 - 4

titik stationer (x₁,y₁) dan (x₂,y₂)
= { (-2 +√3 , -4 + 2√3) , (-2 - √3), (-4 - 2√3)}